# Counterexamples of Arzèla Ascoli theorem for non-obeyed criteria

I had an exam on functional analysis some time ago, and one of the questions I couldn't make any sense out was the following:

Let $\Omega\subset \mathbb{R}$ and $\{f_n\}$ a sequence of continuous functions from $\Omega$ to $\mathbb{R}$. If the following criteria are obeyed:

1. $\exists M>0$ such that $||f_n||_{\infty}< M$ $\forall n\in\mathbb{N}$,
2. $\Omega$ is compact,
3. The sequence $\{f_n\}$ is uniform equicontinuous.

then the theorem of Arzèla Ascoli states that the sequence $\{f_n\}$ has a subsequence which converges in the $||.||_{\infty}$norm to a continuous function.

Show that the theorem is not true by stating counter examples in the cases:

1. (1) and (2) are obeyed, but not (3),
2. (1) and (3) are obeyed, but not (2),
3. (2) and (3) are obeyed, but not (1).

I spend a lot of time thinking about this, but I couldn't think of any counter examples. For a non-compact subset of $\mathbb{R}$ I tried $(0,1)$ (as it is not closed) and for a bounded sequence I was thinking of $f_n = x^n$ but these didn't work. Can anyone help me with some counter examples and maybe a good way of thinking of them?

1. Let $\Omega=[0,1]$, $f_n(x)=x^n$. Then $\|f_n\|=1$ for all $n$. But the functions are not equicontinuous. If they were, there would exist $\delta>0$ such that $\delta<1$, for all $x,y\in(1-\delta,1]$, and for all $n$, $$|y^n-x^n|<1/2.$$ But if we take $n$ such that $(1-\delta/2)^n<1/2$, $y=1$ and $x=1-\delta/2$, then $$|y^n-x^n|=1-(1-\delta/2)^n>1-1/2>1/2,$$ a contradiction. So the family is not equicontinuous.

2. Let $\Omega=\mathbb R$, and define $$f_n(x)=\begin{cases}0,&\mbox{ if } x\not\in[n,n+1] \\ 2(x-n),&\mbox{ if } x\in[n,n+1/2]\\ 2-2(x-n),&\mbox{ if } x\in[n+1/2,n+1] \end{cases}.$$ Then $\|f_n\|\leq1$ for all $n$. Given $\varepsilon>0$, take $\delta=\varepsilon/2$. Then $$|f_n(x)-f_n(y)|<\varepsilon$$ whenever $|x-y|<\delta$. So the family is equicontinuous. But $\|f_n-f_m\|_\infty=1$ if $n\ne m$.

3. Take $\Omega=[0,1]$, $f_n(x)=n$. Then the family is equicontinuous (any $\delta$ works for any $\varepsilon$!), but $\|f_n-f_m\|_\infty=|n-m|$.

For your second point, $(0,1)$ will work, but you'll see that any sequence of functions has a subsequence converging uniformly on any compact $K \subset (0,1)$; therefore what you'll be looking for is some nonuniformity at the endpoints $\{0,1\}$.

An easier counterexample is to set $\Omega = \mathbb{R}$ and then use the sequence

$$f_n(x) = \begin{cases}0 & x \leq n \\ x - n & n < x \leq n+1 \\ 1 & x > n+1 \end{cases}$$ The sequence converges pointwise to the zero function in the limit as $n \rightarrow \infty$, converges uniformly on compacts $K \subset \mathbb{R}$, but converges strictly nonuniformly on all of $\mathbb{R}$.

• I just noticed: the sequence $\{x^n\}$ converges pointwise but nonuniformly to zero on $(0,1)$. To see that, consider the convergence properties near $x= 1$. – A Blumenthal Mar 11 '13 at 21:53
• I don't think that Part 2 works for any bounded subset, like $(0,1)$, because the "usual" extension of a family of uniform equicontinuous functions to $[0,1]$ remains uniform equicontinuous. – QA Ngô Apr 19 '19 at 4:11